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Journal of Integer Sequences, Vol. 17 (2014),
Article 14.8.7
23 11
On the Dirichlet Convolution of Completely
Additive Functions
Isao Kiuchi and Makoto Minamide
Department of Mathematical Sciences
Yamaguchi University
Yamaguchi 753-8512
Japan
kiuchi@yamaguchi-u.ac.jp
minamide@yamaguchi-u.ac.jp
Abstract
Let k and l be non-negative integers. For two completely additive functions f and
g, we consider various identities for the Dirichlet convolution of the kth powers of f
and the lth powers of g. Furthermore, we derive some asymptotic formulas for sums
of convolutions on the natural logarithms.
1
Statements of results
Let f and g be two arithmetical functions that are completely additive. That is, these
functions satisfy f (mn) = f (m) + f (n) and g(mn) = g(m) + g(n) for all positive integers m
and n. We shall consider the arithmetical function
X
k
l n
Dk,l (n; f, g) :=
f (d)g
,
(1)
d
d|n
which represents the Dirichlet convolution of the kth power of f and the lth power of g for
non-negative integers k and l. The above function provides a certain generalization of the
classical number-of-divisors function d(n). In fact,
D0,0 (n; f, g) = d(n).
1
The first purpose of this study is to investigate some recurrence formulas for Dk,l (n; f, g)
with respect to k and l. Since
X
d|n
1
f (d) = d(n)f (n),
2
(2)
where f is a completely additive function, we have
X
1
D1,1 (n; f, g) = d(n)f (n)g(n) −
f (d)g(d).
2
(3)
d|n
Similarly, as in (3), we use (1) for Dk,l+1 (n; f, g) to obtain
n n
X
g
Dk,l+1 (n; f, g) =
f k (d)g l
d
d
d|n
n X
n
X
= g(n)
f k (d)g l
−
.
f k (d)g(d)g l
d
d
d|n
d|n
Hence, we deduce the following two recurrence formulas.
Theorem 1. Let k and l be non-negative integers, and let f and g be completely additive
functions. Then we have
X
k
l n
g(d) = g(n)Dk,l (n; f, g),
(4)
Dk,l+1 (n; f, g) +
f (d)g
d
d|n
n n
X
gl
= f (n)Dk,l (n; f, g).
(5)
Dk+1,l (n; f, g) +
f k (d)f
d
d
d|n
Now, we put f = g in (4) (or (5)), and set Dk,l (n; f ) := Dk,l (n; f, f ). Then, we deduce
the following corollary.
Corollary 2. Using the same notation given above, we have
Dk,l+1 (n; f ) + Dk+1,l (n; f ) = f (n)Dk,l (n; f ).
(6)
Particularly, if k = l, we have
1
Dk+1,k (n; f ) = Dk,k+1 (n; f ) = f (n)Dk,k (n; f ).
2
(7)
Because the symmetric property Dk,l (n; f ) = Dl,k (n; f ), we only consider the function
Dk,k+j (n, f ) for j = 1, 2, . . ..
2
Example 3. The formulas (6) and (7) imply that
1
Dk,k+2 (n; f ) = f 2 (n)Dk,k (n; f ) − Dk+1,k+1 (n; f ),
2
1 3
3
Dk,k+3 (n; f ) = f (n)Dk,k (n; f ) − f (n)Dk+1,k+1 (n; f ),
2
2
1 4
Dk,k+4 (n; f ) = f (n)Dk,k (n; f ) − 2f 2 (n)Dk+1,k+1 (n; f ) + Dk+2,k+2 (n; f ),
2
5
5
1 5
Dk,k+5 (n; f ) = f (n)Dk,k (n; f ) − f 3 (n)Dk+1,k+1 (n; f ) + f (n)Dk+2,k+2 (n; f ),
2
2
2
1 6
9 2
4
Dk,k+6 (n; f ) = f (n)Dk,k (n; f ) − 3f (n)Dk+1,k+1 (n; f ) + f (n)Dk+2,k+2 (n; f )
2
2
− Dk+3,k+3 (n; f ),
1
7
Dk,k+7 (n; f ) = f 7 (n)Dk,k (n; f ) − f 5 (n)Dk+1,k+1 (n; f ) + 7f 3 (n)Dk+2,k+2 (n; f )
2
2
7
− f (n)Dk+3,k+3 (n; f ),
2
1 8
Dk,k+8 (n; f ) = f (n)Dk,k (n; f ) − 4f 6 (n)Dk+1,k+1 (n; f ) + 10f 4 (n)Dk+2,k+2 (n; f )
2
− 8f 2 (n)Dk+3,k+3 (n; f ) + Dk+4,k+4 (n; f ).
Next, we shall demonstrate that the explicit evaluation of the function Dk,k+m (n; f )
(m = 2, 3, . . .) can be expressed as a combination of the functions Dk,k (n; f ), Dk+1,k+1 (n; f ),
Dk+2,k+2 (n; f ), . . ., Dk+⌊ m2 ⌋,k+⌊ m2 ⌋ (n; f ). Hence, we shall give a recurrence formula between
Dk,k (n; f ), . . . , Dk+⌊ m2 ⌋,k+⌊ m2 ⌋ (n; f ) and Dk,k+m (n; f ).
Theorem 4. Let k and m be positive integers, and let Dk,k+m (n; f ) be the function defined
by the above formula. Then we have
m
Dk,k+m (n; f ) =
⌊2⌋
X
(m)
ck,j f m−2j (n)Dk+j,k+j (n; f ),
(8)
j=0
where
(m)
ck,j
1

,



2m


− ,
=
2
j−1



m Y

j

(m − (j + i)) ,
(−1) 2 · j!
i=1
if j = 0;
if j = 1;
if 2 ≤ j ≤ ⌊
m
⌋.
2
Proof. By (7) in Corollary 2, the equality (8) holds for m = 1 and all k ∈ N. Now, we
assume that (8) is true for m = 1, 2, . . . , l and k ∈ N. Using this assumption and (6) in
3
Corollary 2, we observe that
l
Dk,k+l+1 (n; f ) =
⌊2⌋
X
(l)
ck,j f l+1−2j (n)Dk+j,k+j (n; f )
j=0
⌋
⌊ l−1
2
X
−
(l−1)
ck+1,j f l−1−2j (n)Dk+1+j,k+1+j (n; f ).
j=0
For even l = 2q, we have
1
Dk,k+2q+1 (n; f ) = f 2q+1 (n)Dk,k (n; f )
2
q X
(2q)
(2q−1)
ck,j − ck+1,j−1 f 2q+1−2j (n)Dk+j,k+j (n; f )
+
j=1
and
(2q)
ck,j
−
(2q−1)
ck+1,j−1
j−2
j 2q
+1Y
(2q+1)
= (−1)
(2q − (j + i)) (2q − j) = ck,j .
2 · j! i=1
For odd l = 2q − 1, we observe that
1
Dk,k+2q (n; f ) = f 2q (n)Dk,k (n; f )
2
q−1 X
(2q−1)
(2q−2)
ck,j
− ck+1,j−1 f 2q−2j (n)Dk+j (n; f )
+
j=1
2q
+ (−1)⌊ 2 ⌋ Dk+⌊ 2q ⌋,k+⌊ 2q ⌋ (n; f ).
2
2
By our assumption, since
j−2
(2q−1)
ck,j
−
(2q−2)
ck+1,j−1
2q Y
(2q)
(2q − 1 − (j + i)) (2q − 1 − j) = ck,j ,
= (−1)
2 · j! i=1
j
we obtain the assertion (8) for all k and m ∈ N.
Now, we consider another expression for Dk,l (n; f, g) using the arithmetical function
X
Hk,m (n; f, g) :=
f k (d)g m (d).
(9)
d|n
4
If f = g, we set Hk+m (n; f ) = Hk,m (n; f, f ). The right-hand side of (9) implies the Dirichlet
convolution of 1 and f k g m . Since g is a completely additive function, we have
X
Dk,l (n; f, g) =
f k (d) (g(n) − g(d))l
d|n
X
=
k
f (d)
l
X
l
g l−m (n)g m (d).
(−1)
m
m
m=0
d|n
From (9) and the above, we obtain the following theorem.
Theorem 5. Let k and l be non-negative integers, and let f and g be completely additive
functions. Then we have
l
X
l
Dk,l (n; f, g) =
g l−m (n)Hk,m (n; f, g),
(−1)
m
m=0
m
where the function Hk,m (n; f, g) is defined by (9).
We immediately obtain the following corollary.
Corollary 6. Let k and l be non-negative integers, and let f and g be completely additive
functions. Then we have
l
X
l
Dk,l (n; f ) =
(−1)
f l−m (n)Hk+m (n; f ).
m
m=0
m
(10)
Note that
Hk,m (n; f, g) =
X
fk
d|n
=
n
k X
m
X
d
gm
(−1)
i=0 j=0
i+j
n
d
X
k
m k−i
f (n)g m−j (n)
f i (d)g j (d).
i
j
(11)
d|n
Applying (11) to Theorem 5, we have the following theorem.
Theorem 7. Let k and l be non-negative integers, and let f and g be completely additive
functions. Then we have
Dk,l (n; f, g)
=
l X
k X
m
X
m=0 i=0 j=0
(−1)
m+i+j
X
l
k
m k−i
f (n)g l−j (n)
f i (d)g j (d).
m
i
j
d|n
5
In the case where f = g, note that
X
Hk+m (n; f ) =
(f (n) − f (d))k+m
d|n
=
k+m
X
(−1)
j=0
j
X
k + m k+m−j
f
(n)
f j (d).
j
d|n
From (10) and the above, we obtain the following corollary.
Corollary 8. Let k and l be non-negative integers, and let f be a completely additive function. Then we have
l k+m
X
X
X
l
k + m k+l−j
m+j
Dk,l (n; f ) =
(−1)
f
(n)
f j (d).
(12)
m
j
m=0 j=0
d|n
2
Recurrence formula connecting Dk,l (n; f ) with
P
d|n f
j
(d)
The second purpose of this study is to derive another expression for Dk,l (n; f ) that involves
the divisor function d(n). Before stating Theorem 10, we prepare the following lemma.
Lemma 9. Let f be a completely additive function. There exist the constants eq,q , eq,q−1 , . . . , eq,1
(q = 1, 2, . . .) that satisfy the equation
X
f
2q−1
(d) = eq,q d(n)f
2q−1
(n) +
q−1
X
eq,q−j f 2q−2j−1 (n)
j=1
d|n
X
f 2j (d).
(13)
d|n
Moreover, the relations among sequences (eq,q−j )qj=1 are as follows.
eq,q
!
q−1 X
(22q − 1) B2q
2q − 1
ej,j =
1−
,
2j
−
1
q
j=1


X
q−1 1  2q − 1
2q − 1

ei,i−j  ,
= 
−
2i − 1
2
2j
i=2
1
=
2
eq,q−j
i−j≥1
where Bn denotes the nth Bernoulli number, which is defined by the Taylor expansion
∞
X Bn
z
=
zn,
ez − 1 n=1 n!
6
(|z| < 2π).
(14)
Proof. By (2), the case q = 1 in (13) is trivial. Assume that there exist ep,p , ep,p−1 , . . . , ep,1
(p ≤ q) such that
X
f
2p−1
(d) = ep,p d(n)f
2p−1
(n) +
p−1
X
ep,p−j f 2p−2j−1 (n)
j=1
d|n
X
f 2j (d).
(15)
d|n
Since
2q+1
X
f
2q+1
(d) =
X
(−1)
j=0
d|n
j
X
2q + 1 2q+1−j
f
(n)
f j (d),
j
d|n
we have
X
d|n
f
2q+1
q X
1 X 2q + 1 2q+1−2j
1
2q+1
f
(n)
(n) +
f 2j (d)
(d) = d(n)f
2j
2
2 j=1
d|n
q X
1 X 2q + 1 2q+2−2j
f
(n)
−
f 2j−1 (d).
2 j=1 2j − 1
(16)
d|n
Applying (15) to (16), we obtain
X
d|n
!
q X
1
2q + 1
f 2q+1 (d) =
ej,j d(n)f 2q+1 (n)
1−
2j − 1
2
j=1


q
q
X
X 2q + 1
1 X  2q + 1

+
ei,i−j  f 2q−2j+1 (n)
−
f 2j (d).

2i
−
1
2 j=1
2j
i=2
d|n
i−j≥1
By induction, this completes the proof, except for the second term on the right-hand side of
(14).
The first term on the right-hand side of (14) implies
eq,q = 1 −
q X
2q − 1
k=1
2k − 1
ek,k = 1 −
q X
2q k
k=1
2k q
ek,k .
Here we put a(k) = kek,k . Then we have
q X
2q
a(k).
a(q) = q −
2k
k=1
(17)
Since a(1) = e1,1 = 1/2 and (22 − 1) B2 = 1/2, we only need to show that 22k − 1 B2k
(k = 1, . . . , q) satisfies the recurrence formula (17). Consider the nth Bernoulli polynomial
7
Bn (x), which is defined by the following Taylor expansion:
∞
X Bn (x)
zexz
=
zn
ez − 1 n=0 n!
(|z| < 2π).
The following relations are known among Bn (1), Bn (1/2) and Bn ,
1
Bn (1) = Bn ,
Bn
= − 1 − 21−n Bn .
2
By the formula [1, Thm. 12.12, p. 264]
Bn (y) =
n X
n
k
k=0
we observe that
2q
B2q (y) = y − qy
2q−1
+
Bk y n−k ,
2q X
2q
k=2
k
(18)
Bk y 2q−k .
In this equation, we consider y = 1 and y = 1/2; then
q X
2q
B2k
B2q = 1 − q +
2k
k=1
(19)
and
1
2 B2q
= 2 − 22q B2q
2
q X
2q
B2k 22k .
= 1 − 2q +
2k
k=1
2q
(20)
Subtracting (20) from (19), we obtain
2q
2 − 1 B2q = q −
q X
2q
k=1
2k
22k − 1 B2k .
This recurrence formula for 22k − 1 B2k ’s is equivalent to (17). This completes the proof
of (13).
Applying Lemma 9 to (12) in Corollary 8, we have
k+m
⌋
⌊X
2
X
k + m l+k−2j
l
f
(n)
Dk,l (n; f ) =
f 2j (d)
(−1)
2j
m j=0
m=0
l
X
l
X
m
d|n
⌊ k+m+1
⌋
2
X X
l
k + m l+k+1−2j
−
(−1)
f
(n)
f 2j−1 (d).
m
2j − 1
m=0
j=1
m
d|n
8
(21)
The second term on the right-hand side of (21) gives us


k+m
⌋
⌊X
l
2
X
l
k + m  l+k
(−1)m
−
ej,j
f (n)d(n)
m
2j
−
1
m=0
j=1
k+m+1
−
⌋ j−1
l ⌊ X
2
X
X
m=0
X
l
k+m
ej,j−i f l+k−2i (n)
(−1)
f 2i (d)
m
2j
−
1
i=1
j=1
m
(22)
d|n
using (13). From (14), (21) and (22), we have the following theorem.
Theorem 10. Let k and l be non-negative integers, and let f be a completely additive
function. There exist the constants ej,j , ej,j−i (j = 1, 2, . . . , ⌊ k+m+1
⌋, 1 ≤ j − i < j) and Ak,l
2
such that
Dk,l (n; f ) =Ak,l f l+k (n)d(n)
k+m
+
⌋
l ⌊X
2
X
m=0 j=0
k+m
−
X
l
k + m l+k−2j
f
(n)
(−1)
f 2j (d)
m
2j
m
⌋ j−1
l ⌊X
2
X
X
m=0 j=1
(23)
d|n
X
k+m
l
ej,j−i f l+k−2i (n)
f 2i (d),
(−1)
2j
−
1
m
i=1
m
d|n
where
Ak,l
⌋
⌊ k+m+1
2
X
l
(22j − 1)B2j k + m
m−1
=
(−1)
j
2j − 1
m
m=0
j=1
k+m+1
l
X
−1
l 2
m
=2
(−1)
Bk+m+1 .
m
k
+
m
+
1
m=0
l
X
Proof. We only need to show (24) to complete the proof of Theorem 10. We set
Ak,l
⌊ k+m+1
⌋
2
l
X
l
m−1
=
Jk,m ,
(−1)
m
m=0
From the identity
1 k+m
j 2j−1
Jk,m
=
Jk,m =
j=1
k+m+1
2
2j
k+m+1
2
=
k+m+1
X
(22j − 1)B2j k + m
.
j
2j − 1
, we have
⌋
⌊ k+m+1
2
X
j=1
k+m+1
(2 − 1)B2j
.
2j
2j
9
(24)
Since B2j+1 = 0 if j ≥ 1 and B1 = − 21 , we have
Jk,m
Taking y =
1
2
k+m+1
X
k+m+1
2
2j
(2 − 1)Bj
+ 1.
=
k + m + 1 j=0
j
(25)
and y = 1 in (18), we get
X
n
1
j n
Bj
2 Bn
=
2
j
2
j=0
n
and
Bn =
n X
n
j=0
j
Bj ,
respectively. Hence we have
n
X
n
1
j
2 Bn
Bj .
− Bn =
2 −1
j
2
j=0
n
(26)
On the other hand, it is known that
Bn (mx) = m
n−1
m−1
X
j=0
j
.
Bn x +
m
See [1, p. 275]. Taking x = 0 and m = 2 in (27), we obtain
1
n
2 Bn
− Bn = − (2n − 1) Bn .
2
Then we have, from (26) and the above,
n
− (2 − 1) Bn =
n
X
j=0
n
.
2 − 1 Bj
j
j
Substituting this relation in (25), we have
Jk,m = −
2
(2k+m+1 − 1)Bk+m+1 + 1.
k+m+1
Hence we have
Ak,l
l
X
k+m+1
l 2
−1
Bk+m+1 .
=2
(−1)
k
+
m
+
1
m
m=0
m
From (23), we obtain the following corollary.
10
(27)
Corollary 11. For every positive integer k, there are constants ck , ck−1 , . . . , c0 such that
2k
Dk,k (n; f ) = ck d(n)f (n) +
k
X
ck−j f 2k−2j (n)
j=1
X
f 2j (d).
(28)
d|n
From (8) and (28), we obtain the following corollary.
Corollary 12. For every positive integer k and m (≥ 2), there are constants c0 , c1 , . . . , ck ,
(m)
(m)
(m)
(m)
. . . , ck+⌊ m2 ⌋ as in Corollary 11 and ck,1 = − m2 , ck,2 = m(m−3)
, ck,3 , . . ., ck,⌊ m ⌋ as in Theorem
4
2
4 such that


⌋
⌊m
2
X
1
(m)
Dk,k+m (n; f ) =  ck +
ck,p ck+p  d(n)f 2k+m (n)
2
p=1
k
X
1X
+
ck−j f 2k+m−2j (n)
f 2j (d)
2 j=1
d|n
⌋
⌊m
2
+
X
p=1
(m)
ck,p
k+p
X
ck+p−j f 2k+m−2j (n)
j=1
X
f 2j (d).
d|n
Example 13. Corollaries 11 and 12 give us
X
1
f 2 (d),
D1,1 (n; f ) = d(n)f 2 (n) −
2
d|n
X
1
1
D2,1 (n; f ) = d(n)f 3 (n) − f (n)
f 2 (d),
4
2
d|n
X
X
3
1
f 2 (d) −
f 4 (d),
D3,1 (n; f ) = − d(n)f 4 (n) + f 2 (n)
4
2
d|n
1
D2,2 (n; f ) = d(n)f 4 (n) − 2f 2 (n)
2
X
1
D3,2 (n; f ) = d(n)f 5 (d) − f 2 (n)
4
X
d|n
11
f 2 (d) +
d|n
5
1
D4,1 (n; f ) = − d(n)f 5 (n) + f 3 (n)
2
2
d|n
X
d|n
X
f 4 (d),
d|n
X
3
f 2 (d) − f (n)
f 4 (d),
2
1
f 2 (d) + f (n)
2
d|n
X
d|n
f 4 (d).
3
Applications
The second author [2, p. 330] showed an asymptotic formula for
P
n≤x
D1,1 (n; log)
1
1
D1,1 (n; log) = x log3 x − x log2 x + (1 − 2A1 )x log x
6
2
n≤x
1 + (2A1 − 4A2 − 1)x + Oε x 3 +ε
X
(29)
for x > 2 and all ε > 0. Here the constants A1 and A2 are coefficients of the Laurent
expansion of the Riemann zeta-function ζ(s) in the neighbourhood s = 1:
ζ(s) =
1
+ A0 + A1 (s − 1) + A2 (s − 1)2 + A3 (s − 1)3 + · · · .
s−1
We use (7), (29) and Abel’s identity [1, Thm. 4.2, p. 77] to obtain
X
D2,1 (n; log) =
n≤x
1
1
x log4 x − x log3 x + (1 + A1 )x log2 x
12
3
1 − 2(1 + A2 )x log x + 2(1 + A2 )x + Oε x 3 +ε .
Furthermore, a generalization of (29) for the partial sums of Dk,k (n; log) for positive
integers k was considered by the second author [2, Thm. 1.2, p. 326], who demonstrated
that there exists a polynomial P2k+1 of degree 2k + 1 such that
1 X
Dk,k (n; log) = xP2k+1 (log x) + Ok,ε x 3 +ε
(30)
n≤x
for every ε > 0. Applying Theorem 4 and the above formula (30), we have the following
theorem.
Theorem 14. There exists a polynomial U2k+m+1 of degree 2k + m + 1 such that
1 X
Dk,k+m (n; log) = xU2k+m+1 (log x) + Ok,m,ε x 3 +ε
n≤x
for m = 2, 3, . . . and every ε > 0.
4
Acknowledgment
The authors deeply thank the referee for carefully reading this paper and indicating some
mistakes.
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References
[1] T. M. Apostol. Introduction to Analytic Number Theory. Springer-Verlag, 1976.
[2] M. Minamide. The truncated Voronoı̈ formula for the derivative of the Riemann zetafunction. Indian J. Math. 55 (2013), 325–352.
2010 Mathematics Subject Classification: Primary 11A25; Secondary 11P99.
Keywords: completely additive function, Dirichlet convolution.
Received April 23 2014; revised versions received May 15 2014; July 9 2014; July 20 2014;
July 31 2014. Published in Journal of Integer Sequences, August 5 2014.
Return to Journal of Integer Sequences home page.
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